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Each of the 25 balls in a certain box is either red, blue or

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Each of the 25 balls in a certain box is either red, blue or [#permalink] New post 13 Aug 2010, 02:03
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Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

(1) The probability that the ball will both be white and have an even number painted on it is 0

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2
[Reveal] Spoiler: OA
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Re: GMAT prep DS [#permalink] New post 13 Aug 2010, 02:42
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Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W);
Probability ball: even - P(E);
Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.
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Re: GMAT prep DS [#permalink] New post 13 Aug 2010, 06:59
But how can this be solved in less than 2 mins???
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Re: GMAT prep DS [#permalink] New post 13 Aug 2010, 08:27
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jananijayakumar wrote:
But how can this be solved in less than 2 mins???


You can solve this problem in another way. Transform probability into actual numbers and draw the table.

Given:
Attachment:
1.JPG
1.JPG [ 8.64 KiB | Viewed 4587 times ]


So we are asked to calculate \frac{a+b-c}{25} (we are subtracting c not to count twice even balls which are white).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> c=0 --> \frac{a+b}{25}=?. Not sufficient.
Attachment:
4.JPG
4.JPG [ 8.7 KiB | Viewed 4590 times ]


(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> \frac{white}{25}-\frac{even}{25}=0.2 --> white-even=25*0.2=5 --> a-b=5 --> b=a-5 --> \frac{a+a-5-c}{25}=?. Not sufficient.
Attachment:
2.JPG
2.JPG [ 8.68 KiB | Viewed 4586 times ]


(1)+(2) c=0 and b=a-5 --> \frac{a+a-5+0}{25}=\frac{2a-5}{25}. Not sufficient.
Attachment:
3.JPG
3.JPG [ 8.82 KiB | Viewed 4588 times ]

Answer: E.
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Re: GMAT prep DS [#permalink] New post 14 Aug 2010, 08:00
great explanation bunuel.
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Re: GMAT prep DS [#permalink] New post 15 Aug 2010, 08:37
Wonderful Bunuel. :-) Thanks so much!
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Re: GMAT prep DS [#permalink] New post 08 Oct 2010, 15:31
Bunuel wrote:
Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W);
Probability ball: even - P(E);
Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.


how did you get this?
P(WorE)=2P(E)+0.2
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Re: GMAT prep DS [#permalink] New post 08 Oct 2010, 15:35
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BlitzHN wrote:
Bunuel wrote:
Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W);
Probability ball: even - P(E);
Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.


how did you get this?
P(WorE)=2P(E)+0.2


From (1) P(WorE)=P(W)+P(E)-0 --> P(WorE)=P(W)+P(E);
From (2) P(W)-P(E)=0.2 --> P(W)=P(E)+0.2;

Substitute P(W) --> P(WorE)=P(W)+P(E)=P(E)+0.2+P(E)=2P(E)+0.2.
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Re: GMAT prep DS [#permalink] New post 08 Oct 2010, 15:37
very cool.. thanks!
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Re: Each of the 25 balls in a certain box is either red, blue or [#permalink] New post 03 Mar 2012, 04:12
Hi
But can we not say that 12 out of the 25 balls were even ?
If we can then we already get the answer with only B !.

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Re: Each of the 25 balls in a certain box is either red, blue or [#permalink] New post 03 Mar 2012, 04:48
Expert's post
morya003 wrote:
Hi
But can we not say that 12 out of the 25 balls were even ?
If we can then we already get the answer with only B !.

Thanks & Regards


Well since OA is E then apparently we can not.

Also how did you get 12? Anyway even if we knew from (2) that # of even balls is 12 the answer still wouldn't be B since we would need # of balls which are both white and have an even number painted on them, so in this case answer would C. Please refer to: each-of-the-25-balls-in-a-certain-box-is-either-red-blue-or-99044.html#p763481
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Re: Each of the 25 balls in a certain box is either red, blue or [#permalink] New post 03 Mar 2012, 06:32
Dont know why I have already put 1000 Dollars behind GMAT - lolz

I calculated 12 even numbers as follows - first 10 balls can be painted with 5 even numbers viz - 2,4,6,8,10
likewise for next 10 - 2,4,6,8,10 and next 5 - 2,4,

then from Statement B

P(W) - P(E) = 0.2
So P (W) = 0.2 + P(E)

P(E) = 12/25

P(W) = 0.2 + 12/25

:lol: :lol: :lol: :lol: :lol: :lol:
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Re: GMAT prep DS [#permalink] New post 03 Jul 2012, 08:30
Bunuel wrote:
Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W);
Probability ball: even - P(E);
Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.


I found this method simple although it might look complicated. Took me exactly 2 minutes.
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Re: GMAT prep DS [#permalink] New post 14 Apr 2013, 17:37
Bunuel wrote:
Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W);
Probability ball: even - P(E);
Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.


The interesting part about this explanation is particularly helpful expression "multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE)." I did not think of that at all (I just thought, well this is a minus probabilities and we need plus, so not sufficient- I am not even sure I understood any relevance of the 2nd option), when I finally thought of the problem in the same way. Now, may be it is quite late here and my brain refuses to come up with something :-D , but are there instances in which multiple values are not possible and hence the answer would be B? Sort of a "what if" on this problem...
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Probability [#permalink] New post 22 Apr 2013, 23:47
Each of the 25 balls in acertain box is red, blue or white and has a number from 1 to 10 painted on it. If one ball is selected at random from the box , what is the probability that the ball selected will either be white or have an even number painted on it ?

a. the probability that the ball selected will be white and an even number painted on it is 0
b.The probability that the ball will be white minus an even number painted on it is 0.2
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Re: Probability [#permalink] New post 23 Apr 2013, 04:39
Expert's post
kabilank87 wrote:
Each of the 25 balls in acertain box is red, blue or white and has a number from 1 to 10 painted on it. If one ball is selected at random from the box , what is the probability that the ball selected will either be white or have an even number painted on it ?

a. the probability that the ball selected will be white and an even number painted on it is 0
b.The probability that the ball will be white minus an even number painted on it is 0.2


Merging similar topics.
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Re: GMAT prep DS [#permalink] New post 04 Aug 2013, 05:05
Bunuel wrote:
Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W);
Probability ball: even - P(E);
Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.


You say in the second statement that multiple values are possible (0.6 and 0.4 or 0.4 and 0.2)
Are these values only possible? If they are please explain why.
Why they cannot be 0.3 and 0.1 or 0.5 and 0.3?
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Re: GMAT prep DS [#permalink] New post 05 Aug 2013, 09:54
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Stiv wrote:
Bunuel wrote:
Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W);
Probability ball: even - P(E);
Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.


You say in the second statement that multiple values are possible (0.6 and 0.4 or 0.4 and 0.2)
Are these values only possible? If they are please explain why.
Why they cannot be 0.3 and 0.1 or 0.5 and 0.3?




solution:

White balls = w
Red = R
Blue = b
Total ball = 25
Sum total of even numbered ball = E

We have to evaluate = w/25 + (E)/25

From st(1) , we only know there is no white ball which contains even number. Even we still don’t know about red and blue balls have how many even numbered balls in them. So all are in mystery and doubt.

From st(2), w/25 – E/25 = 0.2 , again all unknown.

From both statement, we can assume several different things, I mean double case.
So both are insufficient. Answer is (E)
_________________

Asif vai.....

Re: GMAT prep DS   [#permalink] 05 Aug 2013, 09:54
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